CODEWITHSUNDEEP
typescriptspread-syntax
paroxyzm
paroxyzm

Reputation: 1499

Spread operator not type safe

I have a function that is supposed to return some type but using the spread operator caused it to assign a misspelled key.

interface State {
    fieldA: number,
    fieldB: string
}

const reducer: (state: State, action: {payload: string}) => State = (state, action) => {

    // please note that those variables are not desired
    const tmp = { ...state, feildB: action.payload }; // No compile time error, :(

    // This is too verbose... but works
    const tmp2 = Object.assign<State, Partial<State>>(state, {feildB: action.payload}) // ERROR - this is what I need
    return tmp
}
const t = reducer({fieldA: 1, fieldB: 'OK'}, {payload: 'Misspelled'}) // Misspelled
console.log("feildB", (t as any).feildB) // Misspelled
console.log("fieldB", (t as any).fieldB) // OK

Is there a way to make it typesafe, while keeping the boilerplate to the minimum?

Playground code HERE

Upvotes: 4

Views: 2108

Answers (1)

Kousha
Kousha

Reputation: 36299

TypeScript is doing what it is supposed to do. In your case, you are creating a new object tmp with a new type that has 3 fields, namely:

interface State {
    fieldA: number;
    fieldB: string;
}
interface Tmp {
    fieldA: string;
    fieldB: string;
    payload: string;
}

In other words, the spread operator performs the following operation:

interface Obj {
    [key: string]: any;
}

const spread = (...objects: Obj[]) => {
    const merged: Obj = {};

    objects.forEach(obj => {
        Object.keys(obj).forEach(k => merged[k] = obj[k]);
    });

    return merged;
}

The spread operator is creating a new type of object for you; if you want to infer the type, then you should do:

// this now throws an error
const tmp: State = { ...state, feildB: action.payload };

Upvotes: 5

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